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A stone of 1 kg is thrown witha velocity of 20m/s -¹ accross the frozen surface of a lake and comes to rest after travelling a distance of 50m. What is the force of friction between stone and ice ?



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Answer 1

Answer:

-4N

Explanation:

Given that,

A stone of 1kg is thrown with a velocity of $\rm 20m/s^{-1}$ across a frozen surface of a lake.

After traveling a distance of 50m it comes to rest

Find the force of friction between stone and ice.

Force of friction is given by,

→ f = m*a

Where,

• m denotes the mass of the stone.
• a is the acceleration.

By equation of motion,

→ v² = u² + 2as

Where,

• v(final velocity) = 0 (since it comes to rest)
• u(initial velocity) = $\rm 20m/s^{-1}$
• s(distance) = 50m

Then,

→ (0)² = (20)² + 2.a.(50)

→ 0 = 400 + 100a

→ -400 = 100a

→ -400/100 = a

→ -4 = a

∴ Acceleration of the stone is -4m/s²

Hence, the frictional force acting is :-

→ f = m*a

→ f = 1*(-4)

→ f = -4

∴ Friction force between ice and stone is -4N

Answer 2

Given :-

A stone of 1 kg is thrown with a velocity of 20m/s -¹ across the frozen surface of a lake and comes to rest after traveling a distance of 50m.

To Find :-

Force of friction between stone and ice

Solution :-

Since the stone is stopped by covering a certain distance of 50 m. And the initial velocity of the stone is given as 20 m/s. By using the third equation of kinematics

v² - u² = 2as

Here

v is the final velocity

u is theinitial velocity

a is the acceleration

s is the distance

(0)² - (20)² = 2(a)(50)

0 - 400 = 100a

-400 = 100a

-400/100 = a

-4/1 = a

-4 = a

Hence, acceleration is -4 m/s²

Now

Force is the product of mass and acceleration

F = m × a

here

f is the force

m is themass

a is the acceleration

F = 1 × -4

F = -4 N

Note = -ve sign shows that the force is applied on opposite direction